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The projection of the vector 2i^−j^+3k^2\hat{i} - \hat{j} + 3\hat{k} on the vector 3i^+2j^+6k^3\hat{i} + 2\hat{j} + 6\hat{k} is

Solution

✅ Correct Option: 1

The projection of vector a⃗\vec{a} on vector b⃗\vec{b} is given by:

Projection =a⃗⋅b⃗∣b⃗∣= \dfrac{\vec{a} \cdot \vec{b}}{|\vec{b}|}

where a⃗=2i^−j^+3k^\vec{a} = 2\hat{i} - \hat{j} + 3\hat{k} and b⃗=3i^+2j^+6k^\vec{b} = 3\hat{i} + 2\hat{j} + 6\hat{k}.


The dot product a⃗⋅b⃗\vec{a} \cdot \vec{b}:

a⃗⋅b⃗=(2)(3)+(−1)(2)+(3)(6)\vec{a} \cdot \vec{b} = (2)(3) + (-1)(2) + (3)(6)

=6−2+18= 6 - 2 + 18

=22= 22


The magnitude of b⃗\vec{b}:

∣b⃗∣=(3)2+(2)2+(6)2|\vec{b}| = \sqrt{(3)^2 + (2)^2 + (6)^2}

=9+4+36= \sqrt{9 + 4 + 36}

=49= \sqrt{49}

=7= 7


The projection:

Projection =a⃗⋅b⃗∣b⃗∣= \dfrac{\vec{a} \cdot \vec{b}}{|\vec{b}|}

=227= \dfrac{22}{7}

Therefore, the projection of the vector 2i^−j^+3k^2\hat{i} - \hat{j} + 3\hat{k} on the vector 3i^+2j^+6k^3\hat{i} + 2\hat{j} + 6\hat{k} is 227\dfrac{22}{7}.

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